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Solution Manual for Introduction to Flight 8th Edition by Anderson

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SOLUTIONS MANUAL TO ACCOMPANY INTRODUCTION TO FLIGHT 8th Edition By John D. Anderson, Jr. Chapter 2 2.1  = p/RT = (1.2)(1.01105 )/(287)(300)  = 1.41 kg/m2 v = 1/ = ... 1/1.41= 0.71 m3/kg 2.2 Mean kinetic energy of each atom = 3 3 23 20 k T = 2 2 (1.38  10 ) (500) = 1.035  10 J One kg-mole, which has a mass of 4 kg, has 6.02 × 1026 atoms. Hence 1 kg has 1 (6.02  1026 ) = 1.505  1026 4 atoms. Total internal energy = (energy per atom)(number of atoms) = (1.035´ 10- 20)(1.505 ´ 1026) = 1.558 ´ 106 J 2.3  = p = 2116 = 0.00237 slug RT (1716)(460 + 59) ft3 Volume of the room = (20)(15)(8) = 2400 ft3 Total mass in the room = (2400)(0.00237) = 5.688slug Weight = (5.688)(32.2) = 183lb 2.4  = p RT = 2116 (1716)(460 - = 0.00274 slug 10) ft3 Since the volume of the room is the same, we can simply compare densities between the two problems.  = 0.00274 - % change =   0.00237 = 0.00037 slug ft3 = 0.00037 ´ (100) = 15.6% increase 0.00237 2.5 First, calculate the density from the known mass and volume,  = 1500/ 900 = 1.67 lbm /ft3 In consistent units,  = 1.67/32.2 = 0.052 slug/ft3. Also, T = 70 F = 70 + 460 = 530 R. Hence, p = RT = (0.52)(1716)(530) p = 47, 290 lb/ft2 or p = 47, 290 / 2116 = 22.3 atm 2.6 p = RT l np = l np + l nR + l nT Differentiating with respect to time, 1 dp = 1 d + 1 dT p dt  dt T dt or, dp = p d + p dT dt  dt T dt or, dp = RT d + R dT dt dt dt (1) At the instant there is 1000 lbm of air in the tank, the density is  = 1000 / 900 = 1.11lb /ft3  = 1.11/32.2 = 0.0345slug/ft3 Also, in consistent units, is given that T = 50 + 460 = 510 R and that dT = 1F/min = 1R/min = 0.016R/sec dt From the given pumping rate, and the fact that the volume of the tank is 900 ft3, we ............................................Continued.......................................... [Show More]

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